Solutions for Questions 4 (page ).

Solution 3.1:

Here are the answers, which you really should check by differentiation. The more differentiation you do the better a person you will become.

$${\textstyle\frac{1}{27}}$$(3x - 1)⁹,     - $${\textstyle\frac{1}{4}}$$e^{1 - 4x},     - $${\textstyle\frac{1}{3}}$$cos x³,    $${\textstyle\frac{1}{6}}$$(2x² - 1)^3/2,    $${\textstyle\frac{1}{2}}$$arcsin(x) + $${\textstyle\frac{1}{2}}$$x$$\sqrt{1-x^2}$$

$$\sqrt{x^2+1}$$,    $${\textstyle\frac{1}{7}}$$sin⁷(x),    $${\textstyle\frac{1}{5}}$$tan⁵(x)

Solution 3.2:

The substitutions that I used are given in brackets after the answer. There is nothing wrong in choosing different ones, provided that they give the answer.

$${\textstyle\frac{1}{40}}$$(5x - 3)⁸,  (y = 5x - 3),    $${\textstyle\frac{1}{6}}$$e^{6x - 7},  (y = 6x - 7),    $${\textstyle\frac{1}{4}}$$sin x⁴,  (y = x⁴)

- $$\sqrt{1-x^2}$$,  (y = 1 - x²),    $${\textstyle\frac{2}{9}}$$(x³ - 1)^3/2,  (y = x³ - 1),     - $${\frac{1}{3\sin^3(x)}}$$,  (y = sin x)

- $${\textstyle\frac{1}{2}}$$e^-x2,  (y = x²),     - $${\textstyle\frac{1}{5}}$$cos⁵(x),  (y = cos x),     - $${\textstyle\frac{1}{10}}$$cos⁵(2x),  (y = cos 2x)

$$\sqrt{x^2+2x+3}$$,  (y = x² + 2x + 3),    $$\sqrt{2x-1}$$,  (y = 2x - 1),    $${\textstyle\frac{1}{2}}$$ln²t,  (y = ln t)

Solution 3.3: Just do the integrals, but note that $\int$sin(a - b)x dx = - cos(a - b)x/(a - b) does not make sense if a = b.

Solution 3.4:Putting y = x - T we get $\int_{T}^{2T}$f (x) dx = $\int_{0}^{T}$f (y + T) dy which, by periodicity, is $\int_{0}^{T}$f (y) dy. Think a bit about the next part. It is not a straightforward substitution. You have to chop up the range in clever ways. Pay attention to where a multiple of T comes in the range [a, a + T].

Solution 3.5:No solution -- check your answers by differentiating!